0 votes 0 votes Consider a matrix $\left[\begin{array}{lll}0 & 1 & 0 \\ a & 2 & d \\ b & 3 & c\end{array}\right]$. The matrix cannot have rank. $0$ $1$ $2$ $3$ Others gateda-sample-paper-2024 + – admin asked Oct 21, 2023 • edited Oct 24, 2023 by makhdoom ghaya admin 1.7k views answer comment Share Follow See 1 comment See all 1 1 comment reply Akash 15 commented Feb 1 reply Follow Share $0$ rank possible only for $Zero\; matrix$ option $A$ 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes Rank of a matrix: The rank of a matrix represents the maximum number of linearly independent rows or columns in the matrix. Zero rank: A matrix has rank 0 only if all its elements are zero. In the given matrix, even when a = b = c = 0, the second row still has a non-zero element (2). Therefore, it cannot have rank 0, as it has at least one linearly independent row. Rank 1: If all rows are linearly dependent (e.g., a = 0, b = 1, c = 2), the matrix will have rank 1. Rank 2: If there are two linearly independent rows (e.g., a = 1, b = 0, c = 1), the matrix will have rank 2. Rank 3: If all three rows are linearly independent (e.g., a = 1, b = 2, c = 3), the matrix will have rank 3. rajveer43 answered Jan 5 rajveer43 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes using row operations to get $$\begin{array} {|r|r|}\hline 0 & 1 & 0 \\ \hline a & 0 & d \\ \hline b & 0 & c \\ \hline \end{array}$$ now put a=b=c=d=0 we get rank=1 $$\begin{array} {|r|r|}\hline 0 & 1 & 0 \\ \hline 0 & 0 & 0 \\ \hline 0 & 0 & 0 \\ \hline \end{array}$$ put a=d=c=0 and b $\ne$ 0 we get rank=2 $$\begin{array} {|r|r|}\hline 0 & 1 & 0 \\ \hline 0 & 0 & 0 \\ \hline b & 0 & 0 \\ \hline \end{array}$$ put b = 0 and d =0 and a $\ne$ 0 and c $\ne$ 0 we get rank=3 $$\begin{array} {|r|r|}\hline 0 & 1 & 0 \\ \hline a & 0 & 0 \\ \hline 0 & 0 & c \\ \hline \end{array}$$ proving rank is always atleast 1 converting original matrix to row echelon form r2 = r2 – 2r1 r3 = r3 – 3r1 r3 = r3 – (b/a)r2 r2 = r2 – $\frac{d}{c – (bd)/a}$r3 switch r1 and r2 $$\begin{array} {|r|r|}\hline a & 0 & 0 \\ \hline 0 & 1 & 0\\ \hline 0 & 0 & c - (bd)/a \\ \hline \end{array}$$ as you can see row 2 will always persist no matter what so rank is always at least 1 so ANS – A ssingla answered Oct 28, 2023 • edited Nov 5, 2023 by ssingla ssingla comment Share Follow See all 2 Comments See all 2 2 Comments reply Riya_23 commented Nov 5, 2023 reply Follow Share Why rank can not be 3? What if this is the matrix? 0 votes 0 votes ssingla commented Nov 5, 2023 reply Follow Share you are right i messed up in my calculations , i have edited my answer 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Ans isA( 0) , at any value of a,b,c,d , the value of rank of matrix can't be zero . _sangra_m answered Nov 5, 2023 _sangra_m comment Share Follow See all 0 reply Please log in or register to add a comment.